Question

If R = 12 cm, M = 520 g, and m = 20 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answer 1

Answer:

v = 0.84 m/s

Explanation:

given,

R = 12 cm

M (mass of pulley )= 520 g

m  (mass of block)=  20 g

s = 50 cm = 0.5 m

using conservation of energy

Potential energy = Kinetic energy

 $m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

   $I_{disk}= \dfrac{1}{2}MR^2$  and v = r ω

 $m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{1}{2}MR^2)(\dfrac{v}{R})^2$

 $m g h = \dfrac{1}{2}mv^2 +\dfrac{1}{4}Mv^2$

 $m g h = \dfrac{1}{2}v^2(m +\dfrac{1}{2}M)$

 $v=\sqrt{\dfrac{2mgh}{m + 0.5 M}}$

 $v=\sqrt{\dfrac{2\times 0.020 \times 9.8 \times 0.5}{0.02 + 0.5\times 0.52}}$

      v = √0.7

      v = 0.84 m/s